大数求余（java高精度）

发布时间：2020-12-14 03:00:52 所属栏目：大数据来源：网络整理

导读：Link：http://poj.org/problem?id=2305 Problem： Basic remains Time Limit: ?1000MS ? Memory Limit: ?65536K Total Submissions: ?4933 ? Accepted: ?2081 Description Given a base b and two non-negative base b integers p and m,compute p mod m and

Link：http://poj.org/problem?id=2305

Problem：

Basic remains

Time Limit:?1000MS	?	Memory Limit:?65536K
Total Submissions:?4933	?	Accepted:?2081

Description

Given a base b and two non-negative base b integers p and m,compute p mod m and print the result as a base b integer. p mod m is defined as the smallest non-negative integer k such that p = a*m + k for some integer a.

Input

Input consists of a number of cases. Each case is represented by a line containing three unsigned integers. The first,b,is a decimal number between 2 and 10. The second,p,contains up to 1000 digits between 0 and b-1. The third,m,contains up to 9 digits between 0 and b-1. The last case is followed by a line containing 0.

Output

For each test case,print a line giving p mod m as a base-b integer.

Sample Input

2 1100 101
10 123456789123456789123456789 1000
0

Sample Output

10
789

Source

Waterloo local 2003.09.20

import java.util.*;

import java.io.*; import java.math.*; import java.text.*; public class Main{ public static void main(String args[]) { BigInteger p,ans; int b; Scanner cin=new Scanner(System.in); while(cin.hasNext()) { b=cin.nextInt(); if(b==0) break; p=cin.nextBigInteger(b); m=cin.nextBigInteger(b); ans=p.mod(m); String s; s=ans.toString(b); System.out.println(s); } } }

（编辑：李大同）

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